Question

The probability of the occurence of two independent events (A and B) is known to be reach greater than $\frac{1}{2}$. Its given that probability of the occurence of first event simultaneously with non occurence of the second event is $\frac{3}{25}$. Also the probability of the occurence of second simultaneously with the first occuring is $\frac{8}{25}$ . Find the probability of the events and P(A)+P(B)

• A. 1
• B. 1.2
• C. 1.1
• D. 1.4

Answer 1

Answer: (D)

1.4

It is given that
$P\left(A\right)(1-P(B\left)\right)=\frac{3}{25}$
$P\left(A\right)(1-P(B\left)\right)=\frac{3}{5}.\frac{1}{5}$ ...(i)
And
$P\left(B\right)(1-P(A\left)\right)=\frac{8}{25}$
$P\left(B\right)(1-P(A\left)\right)=\frac{2}{5}.\frac{4}{5}$ ..(ii)
From i if $1-P\left(B\right)=\frac{3}{5}$
Then
$P\left(B\right)=\frac{2}{5}$.
Then from i and ii, we get
$P\left(A\right)=\frac{1}{5}$.
Hence one combination is $(\frac{1}{5},\frac{2}{5})$.
Similarly
If $P\left(B\right)=\frac{4}{5}$
Then we get
$P\left(A\right)=\frac{3}{5}$
Hence
Another pair is $(\frac{3}{5},\frac{4}{5})$
Now it is given that
$P\left(A\right)+P\left(B\right)>\frac{1}{2}$.
Hence the first pair is ruled out.
Thus we are left with
$P\left(A\right)=\frac{3}{5}$
And
$P\left(B\right)=\frac{4}{5}$.
Therefore
$P\left(A\right)+P\left(B\right)$
$=\frac{7}{5}$
$=1.4$.